Consider an LC circuit, with inductance L = 0 . 1H and capacitance...

Emf induced in the circuit is

$\left|E\right| = \left|\dfrac{dϕ}{dt}\right| = \dfrac{d}{dt}\left[\left({B}_{0} + \betat\right)A\right]$

$= \beta \times A$

$= 0 . 04V$

So the circuit can be rearranged as

Using Kirchhoff's law we can write

$E = L\dfrac{di}{dt} + \dfrac{q}{C}$

$L\dfrac{di}{dt} = E - \dfrac{q}{C}$

For maximum current, $\dfrac{d i}{d t} = 0$,

Or $q = CE$

Using work energy theorem,

$qE = \dfrac{1}{2}L{i}^{2} + \dfrac{{q}^{2}}{2C}$

At {i}_\max, CEE = \dfrac{1}{2}L{i}_\max^{2} + \dfrac{{E}^{2}{C}^{2}}{2C}

Or, {i}_\max = \sqrt{\dfrac{C}{L}}E

So, {i}_\max = \sqrt{\dfrac{{10}^{- 3} \times 0 . 04}{0 . 1 \times {10}^{- 3}}}$= 4mA$